Meaning and Need of Dispersion

While measures of central tendency such as mean, median, and mode indicate a representative value of a data set, they do not describe how the observations are spread around that central value. Two different distributions may have the same average but differ widely in their variability. This limitation leads to the concept of dispersion, which studies the extent of scatter or variability of the observations.

A measure of dispersion quantifies how far the individual values of a distribution deviate from a central value. The greater the dispersion, the more scattered the data; the smaller the dispersion, the more clustered the observations.

Characteristics of a Good Measure of Dispersion

A satisfactory measure of dispersion should:

• Be rigidly defined.
• Be simple to understand and easy to compute.
• Be based on all observations, as far as possible.
• Be amenable to algebraic treatment.
• Not be unduly affected by extreme values.

No single measure fulfills all these conditions perfectly; hence, different measures are used for different purposes.

Absolute Measures of Dispersion

Absolute measures express dispersion in the same units as the data. The main absolute measures are:

1. Range
2. Quartile Deviation
3. Mean Deviation
4. Standard Deviation

The range of a distribution is defined as the difference between the largest and the smallest observations. \[\mathrm{Range}=L-S\] where \(\mathrm{L}\) is the largest value and \(\mathrm{S}\) is the smallest value.

Merits

• It is the simplest measure of dispersion.
• It gives a quick idea of the spread.

Limitations

• It depends only on two extreme values.
• It is highly sensitive to extreme observations.
• It does not represent the overall variability accurately.

Coefficient of Range

To compare dispersion of two distributions, the relative measure of range is used: \[\mathrm{Coefficient\ of\ Range}=\frac{L-S}{L+S}\]

Quartile Deviation is defined as half the difference between the third and first quartiles. \[\mathrm{Quartile\ Deviation}=\frac{Q_3-Q_1}{2}\]

It measures the spread of the middle 50% of the data.

Merits

• It is less affected by extreme values.
• It is suitable for open-ended distributions.

Limitations

• It ignores 50% of the observations.
• It lacks algebraic tractability.

Coefficient of Quartile Deviation

\[\mathrm{Coefficient\ of\ Q.D.}=\frac{Q_3-Q_1}{Q_3+Q_1}\]

Mean Deviation is the arithmetic mean of the absolute deviations of observations from a central value (mean or median).

For a set of observations \(x_1,x_2,\ldots,x_n,\) \[\mathrm{Mean\ Deviation\ about\ }A=\frac{1}{n}\sum\Bigm| x_i-A\Bigm|\] where \(\mathrm{A}\) may be the mean or median.

Mean Deviation about Mean \((\overline{x})\)

\[\mathrm{M.D. (\overline{x})}=\frac{1}{n}\sum\Bigm| x_i-\overline{x}\Bigm|\]

Mean Deviation about Median \(\mathrm{(M)}\)

\[\mathrm{M.D. (M)}=\frac{1}{n}\sum\Bigm| x_i-M\Bigm|\]

Discrete frequency distribution

In a discrete frequency distribution, we organize data that consists of \(n\) distinct values—let's call them \(x_1, x_2, \dots, x_n\)—where each value \(x_i\) occurs a specific number of times, known as its frequency \(f_i\).

\(x\)	\(x_1\)	\(x_2\)	\(x_3\)	\(\ldots\)	\(x_n\)
\(f\)	\(f_1\)	\(f_2\)	\(f_3\)	\(\dots\)	\(f_n\)

The table immediately shows both the unique data points across the top row and how often each one appears in the bottom row. This compact representation makes it easy to visualize patterns, calculate measures like the mean \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\), or prepare for further statistical analysis. Unlike continuous data, discrete distributions deal only with countable, separate values, which is why the frequency table becomes such a natural and powerful summary tool.

Mean deviation about mean

First of all we find the mean \(x\) of the given data by using the formula \[\overline{x}=\dfrac{\sum_\limits{i=1}^n x_if_i}{\sum\limits_{i=1}^n x_i}=\dfrac{1}{N}\sum_{i=1}^n x_if_i\] where
\(\sum_\limits{i=1}^n x_if_i\quad:\quad\) Sum of the products of observations \(x_i\) with their respective frequencies \(f_i\)
\(N=\sum\limits_{i=1}^n f_i\quad:\quad\) Sum of frequencies

Next, we calculate the deviations of each observation \(x_i\) from the mean \(\overline{x}\) specifically, the values \(\mid x_i-x\mid\) for all \(i=1,\;2,\;\ldots,\;n\).
Then, we find the average of these absolute deviations, which gives us the mean deviation about the mean. This measure captures the typical spread of the data around the central value in a straightforward way. \[M.D. (\overline{x})=\dfrac{\sum\limits_{i=1}^n f_i\Bigm| x_i-\overline{x}\Bigm|}{\sum\limits_{i=1}^n f_i}=\dfrac{1}{N}\sum_{i=1}^n f_i\Bigm|x_i-\overline{x}\Bigm|\]

Mean deviation about median

To find the mean deviation about the median, first determine the median of the discrete frequency distribution. Arrange the observations in ascending order, compute the cumulative frequencies, and locate the observation where the cumulative frequency first reaches or exceeds \(\frac{N}{2}\), with N being the total sum of all frequencies.
This observation represents the central position in the ordered data, serving as the median.
Once identified, calculate the absolute deviations \(\mid x_i-\mathrm{median}\mid\) for each observation, then take their average to obtain the mean deviation about the median. Thus \[M.D. (M)=\dfrac{1}{N}\sum_{i=1}^n f_i\Bigm|x_i-M\Bigm|\]

Continuous frequency distribution

A continuous frequency distribution is a series in which the data are classified into different class-intervals without gaps alongwith their respective frequencies.

Mean deviation about mean

When calculating the mean for a continuous frequency distribution, we assume the frequency of each class is concentrated at its midpoint. Similarly here, we take each class's midpoint and follow the same process used for discrete frequency distributions to compute the mean deviation.Thus \[M.D. (\overline{x})=\dfrac{1}{N}\sum_{i=1}^n f_i\Bigm|x_i-\overline{x}\Bigm|\]

Shortcut method for calculating mean deviation about mean

The shortcut method simplifies calculating both the mean and mean deviation about the mean, especially for large datasets.

Choose a convenient assumed mean \(a\) near the data's center. For continuous data or widely spaced values, use class width \(h\) and compute scaled deviations \[d_i = \frac{x_i - a}{h}\], where \(x_i\) represents observations or class midpoints.

To find the mean: Calculate \(\sum f_i d_i\), then \[\bar{x} = a + \frac{\sum_{i=1}^n f_i d_i}{N} \times h\]

To find mean deviation about the mean: Use absolute values \(f_i |d_i|\), sum as \(\sum f_i |d_i|\), then mean deviation = \[\frac{\sum f_i |d_i|}{N} \times h\]

This scaling eliminates large numbers, works perfectly for both discrete and continuous distributions, and saves significant calculation time while matching direct method results exactly.

Mean deviation about median

Mean Deviation about Median for continuous frequency distributions follows the same process as for the mean, but we use the median instead when calculating deviations.

First, locate the median class—the class interval where the cumulative frequency first reaches or exceeds \(\frac{N}{2}\), with \(N\) being the total frequency.

Apply the median formula for that class: \[ \text{Median} = l + \frac{\frac{N}{2} - C}{f} \times h \] where
\(l\) = lower boundary of median class,
\(f\) = frequency of median class,
\(h\) = class width, and
\(C\) = cumulative frequency just before the median class.

Once you have the median \(M\), find the absolute deviations of each class midpoint \(x_i\) from \(M\), weight them by their frequencies \(f_i\), and compute: \[ \text{M.D.}(M) = \frac{1}{N} \sum_{i=1}^n f_i |x_i - M| \]

Standard Deviation is defined as the positive square root of the arithmetic mean of the squares of deviations from the arithmetic mean.
For ungrouped data, \[\sigma=\sqrt{\dfrac{1}{N}\sum_{i=1}^n(x_i-\overline{x})^2}\]

Standard deviation of a discrete frequency distribution

\[\sigma=\sqrt{\dfrac{1}{N}\sum_{i=1}^n f_i\left(x_i-\overline{x}\right)^2}\] where, \(N=\sum\limits_{i=1}^nf_i\)

Another formula for standard deviation

\[ \sigma=\dfrac{1}{N}\sqrt{N\sum_{i=1}^n f_i x_i^2-\left(\sum_{i=1}^n f_i x_i\right)^2} \]

Find the mean deviation about the mean for the following data:
6, 7, 10, 12, 13, 4, 8, 12

Solution

The given data consist of the observations 6, 7, 10, 12, 13, 4, 8 and 12. To find the mean deviation about the mean, we first compute the arithmetic mean of the data.

\[ \begin{aligned} \overline{x} &= \frac{6+7+10+12+13+4+8+12}{8} \\ &= \frac{72}{8} \\ &= 9 \end{aligned} \]

Next, we find the deviations of each observation from the mean by subtracting 9 from every value.

\[ \begin{aligned} x-\overline{x} &= 6-9,\,7-9,\,10-9,\,12-9,\,13-9,\,4-9,\,8-9,\,12-9 \\ &= -3,\,-2,\,1,\,3,\,4,\,-5,\,-1,\,3 \end{aligned} \]

Since mean deviation is based on absolute deviations, we now take the absolute values of these deviations so that negative and positive deviations do not cancel each other.

\[ |x-\overline{x}| = 3,\,2,\,1,\,3,\,4,\,5,\,1,\,3 \]

The mean deviation about the mean is obtained by dividing the sum of these absolute deviations by the total number of observations.

\[ \begin{aligned} \text{Mean Deviation about } \overline{x} &= \frac{\sum\limits_{i=1}^{n} |x-\overline{x}|}{N} \\ &= \frac{3+2+1+3+4+5+1+3}{8} \\ &= \frac{22}{8} \\ &= \frac{11}{4} \end{aligned} \]

Thus, the mean deviation about the mean for the given data is \(\frac{11}{4}\).

Find the mean deviation about the mean for the following data :
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5

Solution

The given data consist of 20 observations. To find the mean deviation about the mean, we first calculate the arithmetic mean of the data.

\[ \begin{aligned} \overline{x} &= \frac{\sum\limits_{i=1}^{n} x_i}{N} \\ &= \frac{12+3+18+17+4+9+17+19+20+15+8+17+2+3+16+11+3+1+0+5}{20} \\ &= 10 \end{aligned} \]

We now find the absolute deviations of each observation from the mean. Absolute values are taken so that deviations on either side of the mean contribute positively to the total dispersion.

\[ \begin{aligned} |x-\overline{x}| &= |12-10|,|3-10|,|18-10|,|17-10|,|4-10|,|9-10|,|17-10|,|19-10|,|20-10|,|15-10|, \\ &\quad |8-10|,|17-10|,|2-10|,|3-10|,|16-10|,|11-10|,|3-10|,|1-10|,|0-10|,|5-10| \\ &= 2,7,8,7,6,1,7,9,10,5,2,7,8,7,6,1,7,9,10,5 \end{aligned} \]

The sum of these absolute deviations is obtained as follows.

\[ \begin{aligned} \sum_{i=1}^{n} |x-\overline{x}| &= 2+7+8+7+6+1+7+9+10+5+2+7+8+7+6+1+7+9+10+5 \\ &= 124 \end{aligned} \]

The mean deviation about the mean is the average of these absolute deviations.

\[ \begin{aligned} \text{Mean Deviation about } \overline{x} &= \frac{\sum\limits_{i=1}^{n} |x-\overline{x}|}{N} \\ &= \frac{124}{20} \\ &= 6.2 \end{aligned} \]

Hence, the mean deviation about the mean for the given data is 6.2.

Find the mean deviation about the median for the following data:
3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.

Solution

The given data consist of 11 observations. To find the mean deviation about the median, we first arrange the observations in ascending order so that the median can be located correctly.

\[ \begin{aligned} \text{Ordered data} &: 3,\,3,\,4,\,5,\,7,\,9,\,10,\,12,\,18,\,19,\,21 \\ N &= 11 \end{aligned} \]

Since the number of observations is odd, the median is the value of the \(\frac{N+1}{2}\)th observation.

\[ \begin{aligned} \text{Position of median} &= \frac{11+1}{2} \\ &= 6 \end{aligned} \]

The 6th observation in the ordered data is 9. Hence, the median \(M = 9\).

We now find the absolute deviations of each observation from the median. Absolute values are taken so that deviations on either side of the median are treated equally.

\[ \begin{aligned} \sum_{i=1}^{n} |x_i - M| &= |3-9|+|3-9|+|4-9|+|5-9|+|7-9|+|9-9| \\ &\quad +|10-9|+|12-9|+|18-9|+|19-9|+|21-9| \\ &= 6+6+5+4+2+0+1+3+9+10+12 \\ &= 58 \end{aligned} \]

The mean deviation about the median is obtained by dividing the sum of absolute deviations by the total number of observations.

\[ \begin{aligned} \text{Mean Deviation about } M &= \frac{\sum_{i=1}^{n} |x_i - M|}{N} \\ &= \frac{58}{11} \\ &= 5.27 \end{aligned} \]

Thus, the mean deviation about the median for the given data is 5.27.

Find mean deviation about the mean for the following data:
\[ \begin{array}{|c|c|} \hline x_i&2&5&6&8&10&12\\ \hline f_i&2&8&10&7&8&5\\ \hline \end{array} \]

Solution

The data are given in the form of a discrete frequency distribution. To find the mean deviation about the mean, we first compute the arithmetic mean using the formula \(\overline{x}=\frac{\sum f_i x_i}{\sum f_i}\).

\(x_i\)	\(f_i\)	\(f_i x_i\)	\(|x_i-\overline{x}|\)	\(f_i|x_i-\overline{x}|\)
2	2	4	5.5	11
5	8	40	2.5	20
6	10	60	1.5	15
8	7	56	0.5	3.5
10	8	80	2.5	20
12	5	60	4.5	22.5
	\(\sum f_i = 40\)	\(\sum f_i x_i = 300\)		\(\sum f_i|x_i-\overline{x}| = 92\)

Using the totals obtained from the table, the arithmetic mean is calculated as follows.

\[ \begin{aligned} N &= \sum f_i = 40 \\ \overline{x} &= \frac{\sum\limits_{i=1}^n f_i x_i}{N} \\ &= \frac{300}{40} \\ &= \frac{15}{2}\\ &= 7.5 \end{aligned} \]

After finding the mean, the absolute deviations of each value from the mean are multiplied by their respective frequencies and added. This gives the total absolute deviation as 92.

The mean deviation about the mean is obtained by dividing this total by the number of observations.

\[ \begin{aligned} \text{Mean Deviation about } \overline{x} &= \frac{1}{N}\sum\limits_{i=1}^n f_i|x_i-\overline{x}| \\ &= \frac{92}{40} \\ &= \frac{23}{10} \end{aligned} \]

Hence, the mean deviation about the mean for the given data is \(\frac{23}{10}\).

Find the mean deviation about the median for the following data: \[ \begin{array}{|c|c|} \hline x_i&3&6&9&12&13&15&21&22\\ \hline f_i&3&4&5&2&4&5&4&3\\ \hline \end{array} \]

Solution

The data are given in the form of a discrete frequency distribution. To find the mean deviation about the median, we first determine the median using cumulative frequencies.

\(x_i\)	\(f_i\)	\(c f\)
3	3	3
6	4	7
9	5	12
12	2	14
13	4	18
15	5	23
21	4	27
22	3	30

The total number of observations is obtained by adding all the frequencies.

\[ \begin{aligned} N &= \sum\limits_{i=1}^n f_i \\ &= 30 \end{aligned} \]

Since the total number of observations is even, the median is the mean of the 15th and 16th observations. From the cumulative frequency table, both the 15th and 16th observations correspond to the value 13.

\[ \begin{aligned} M &= \frac{13+13}{2} \\ &= 13 \end{aligned} \]

We now calculate the absolute deviations of each value from the median and multiply them by the corresponding frequencies.

\[ \text{Mean Deviation about } M = \frac{1}{N}\sum\limits_{i=1}^n f_i|x_i-M| \]

\(x_i\)	\(f_i\)	\(|x_i-M|\)	\(f_i|x_i-M|\)
3	3	10	30
6	4	7	28
9	5	4	20
12	2	1	2
13	4	0	0
15	5	2	10
21	4	8	32
22	3	9	27

The sum of the products \(f_i|x_i-M|\) is obtained from the table.

\[ \begin{aligned} \sum\limits_{i=1}^n f_i|x_i-M| &= 149 \end{aligned} \]

The mean deviation about the median is therefore calculated as follows.

\[ \begin{aligned} \text{Mean Deviation about } M &= \frac{1}{N}\sum\limits_{i=1}^n f_i|x_i-M| \\ &= \frac{149}{30} \\ &= 4.97 \end{aligned} \]

Hence, the mean deviation about the median for the given data is 4.97.

Find the mean deviation about the mean for the following data.
\[ \begin{array}{|c|c|} \hline \text{Marks Obtained}&10-20&20-30&30-40&40-50&50-60&60-70&70-80\\ \hline \text{Number of students}&2&3&8&14&8&3&2\\ \hline \end{array} \]

Solution

The given data are in the form of a continuous frequency distribution. To find the mean deviation about the mean, we first represent each class interval by its class mark, which is taken as the mid-point of the class.

Marks Obtained	No. of Students \(f_i\)	\(x_i\)	\(f_i x_i\)	\(|x_i-\overline{x}|\)	\(f_i|x_i-\overline{x}|\)
10–20	2	15	30	30	60
20–30	3	25	75	20	60
30–40	8	35	280	10	80
40–50	14	45	630	0	0
50–60	8	55	440	10	80
60–70	3	65	195	20	60
70–80	2	75	150	30	60
	\(\sum f_i = 40\)		\(\sum f_i x_i = 1800\)		\(\sum f_i|x_i-\overline{x}| = 400\)

Using the values obtained from the table, the arithmetic mean of the distribution is calculated as follows.

\[ \begin{aligned} N &= \sum\limits_{i=1}^n f_i \\\\ \overline{x} &= \frac{1}{N}\sum\limits_{i=1}^n f_i x_i \\ &= \frac{1800}{40} \\ &= 45 \end{aligned} \]

After finding the mean, the absolute deviations of the class marks from the mean are multiplied by the corresponding frequencies and added. The mean deviation about the mean is then obtained by dividing this total by the number of observations.

\[ \begin{aligned} \text{Mean Deviation about } \overline{x} &= \frac{1}{N}\sum\limits_{i=1}^n f_i|x_i-\overline{x}| \\ &= \frac{400}{40} \\ &= 10 \end{aligned} \]

Hence, the mean deviation about the mean for the given data is 10.

Calculate the mean deviation about median for the following data :
\[ \begin{array}{|c|c|} \hline \text{Class}&0-10&10-20&20-30&30-40&40-50&50-60\\ \hline \text{Frequecy}&6&7&15&16&4&2\\ \hline \end{array} \]

Solution

The given data are in the form of a continuous frequency distribution. To find the mean deviation about the median, we first determine the median of the distribution using cumulative frequencies.

Class	\(f_i\)	\(cf\)	\(x_i\)	\(|x_i-M|\)	\(f_i|x_i-M|\)
0–10	6	6	5	23	138
10–20	7	13	15	13	91
20–30	15	28	25	3	45
30–40	16	44	35	7	112
40–50	4	48	45	17	68
50–60	2	50	55	27	54
	\(\sum f_i = 50\)				\(\sum f_i|x_i-M| = 508\)

The total number of observations is 50, so the median lies at the \(\frac{N}{2}\)th position.

\[ \frac{N}{2} = \frac{50}{2} = 25 \]

The cumulative frequency just exceeding 25 corresponds to the class interval 20–30. Hence, 20–30 is the median class.

\[ M = l + \frac{\frac{N}{2}-c}{f}\times h \]

\[ \begin{aligned} l &= 20 \\ c &= 13 \\ f &= 15 \\ h &= 10 \\ N &= 50 \\ M &= 20 + \frac{25-13}{15}\times 10 \\ &= 20 + \frac{12}{15}\times 10 \\ &= 20 + 8 \\ &= 28 \end{aligned} \]

After determining the median, the absolute deviations of the class marks from the median are multiplied by their respective frequencies. The mean deviation about the median is obtained by dividing the total of these products by the total frequency.

\[ \begin{aligned} \text{Mean Deviation about } M &= \frac{1}{N}\sum\limits_{i=1}^n f_i|x_i-M| \\ &= \frac{1}{50}\times 508 \\ &= 10.16 \end{aligned} \]

Hence, the mean deviation about the median for the given data is 10.16.

Find the variance of the following data:
6, 8, 10, 12, 14, 16, 18, 20, 22, 24

Solution

We are given the data:

\(6, 8, 10, 12, 14, 16, 18, 20, 22, 24\)

The variance is given by

\[ \sigma^{2}=\dfrac{1}{N}\sum_{i=1}^{n}(x_i-\overline{x})^{2} \]

To simplify calculations, we use the step-deviation method. Let the assumed mean be \(a=14\) and common difference \(h=2\). Then

\[ d_i=\dfrac{x_i-a}{h} \]

\(x_i\)	\(d_i=\dfrac{x_i-a}{2}\)	\((x_i-\overline{x})\)	\((x_i-\overline{x})^2\)
6	-4	-9	81
8	-3	-7	49
10	-2	-5	25
12	-1	-3	9
14	0	-1	1
16	1	1	1
18	2	3	9
20	3	5	25
22	4	7	49
24	5	9	81
	\(\sum d_i=5\)		\(\sum (x_i-\overline{x})^2=330\)

First, we compute the arithmetic mean using the step-deviation formula

\[ \begin{aligned} \overline{x} &= a+\dfrac{\sum d_i}{N}\times h\\ &=14+\dfrac{5}{10}\times2\\ &=15 \end{aligned} \]

Now substitute in the variance formula

\[ \begin{aligned} \sigma^{2} &=\dfrac{1}{N}\sum_{i=1}^{n}(x_i-\overline{x})^{2}\\ &=\dfrac{1}{10}\times330\\ &=33 \end{aligned} \]

Hence, the standard deviation is

\[ \sigma=\sqrt{33}\approx5.74 \]

Therefore, the variance of the given data is \(33\) and the standard deviation is approximately \(5.74\).

Calculate the mean, variance and standard deviation for the following distribution:

Class	30-40	40-50	50-60	60-70	70-80	80-90	90-100
\(f\)	3	7	12	15	8	3	2

Solution

To calculate the mean, variance, and standard deviation of the given grouped data, the class marks \(x_i\) are first obtained by taking the mid-point of each class interval. Since all class intervals have equal width \(h=10\), the step–deviation method is used. The assumed mean is chosen as \(a=65\), corresponding to the central class \(60\text{–}70\).

Class	\(f_i\)	\(x_i\)	\(d_i=\dfrac{x_i-a}{10}\)	\(f_i d_i\)	\((x_i-\overline{x})\)	\((x_i-\overline{x})^2\)	\(f_i(x_i-\overline{x})^2\)
30–40	3	35	-3	-9	-27	729	2187
40–50	7	45	-2	-14	-17	289	2023
50–60	12	55	-1	-12	-7	49	588
60–70	15	\(\boxed{\bbox[white,5pt]{\color{blue}65}}\)	0	0	3	9	135
70–80	8	75	1	8	13	169	1352
80–90	3	85	2	6	23	529	1587
90–100	2	95	3	6	33	1089	2178
	\(\sum f_i=50\)			\(\sum f_i d_i=-15\)		\(\sum f_i(x_i-\overline{x})^2=10050\)

The mean is obtained using the step–deviation formula

$$ \begin{aligned} \overline{x}&=a+\dfrac{\sum f_i d_i}{\sum f_i}\times h\\ &=65+\dfrac{-15}{50}\times 10\\ &=62 \end{aligned} $$

Thus, the mean of the distribution is \(62\). The variance is calculated by dividing the sum of squared deviations from the mean, weighted by the frequencies, by the total number of observations.

$$ \begin{aligned} \sigma^2&=\dfrac{1}{N}\sum f_i(x_i-\overline{x})^2\\ &=\dfrac{1}{50}\times 10050\\ &=201 \end{aligned} $$

Therefore, the variance of the distribution is \(201\). The standard deviation is the positive square root of the variance.

$$ \begin{aligned} \sigma&=\sqrt{201}\\ &\approx 14.18 \end{aligned} $$

Hence, the mean, variance, and standard deviation of the given distribution are \(62\), \(201\), and approximately \(14.18\) respectively.

Find the standard deviation for the following data :

\(x_i\)	3	8	13	18	23
\(f_i\)	7	10	15	10	6

Solution

\(x_i\)	\(f_i\)	\(f_ix_i\)	\((x_i)^2\)	\(f_ix_i^2\)
3	7	21	9	63
8	10	80	64
13	15	195	169	2535
18	10	180	324	3240
23	6	138	529	3174
	\(\sum f_i=48\)	\(\sum f_ix_i=614\)		\(\sum f_ix_i^2=9652\)

\[\begin{aligned} \sigma&=\dfrac{1}{N}\sqrt{N\sum_{i=1}^n f_ix_i^2-\left(\sum_{i=1}^n f_ix_i\right)^2}\\\\ &=\dfrac{1}{48}\sqrt{48\times 9652-(614)^2}\\\\ &=\dfrac{1}{48}\sqrt{463296-376996}\\\\ &=\dfrac{1}{48}\times 293.77\\ &=6.12 \end{aligned} \]

Therefore, Standard deviation \(\sigma=6.12\)